Let $\{X_j,\,\, j\ge1\}$ be a sequence of independent r.v.'s having the bernoullian d.f. $P(X_j=1)=p$, $P(X_j=0)=1-p$, $\,\,\,\,\,0<p<1.$ see that ($\{X_j,\,\, j\ge1\}\,\, r.v's \,\,\,\,and\,\,\,\, i.i.d)$
If $\{Y_i,\,\, i\ge1\}$ show that $Y_i=X_{i+1}(1-X_i)$. Prove a central limit theorem for $T_{n}=\sum^{n}_{i=1}Y_i$, it's, $$\frac{T_n-np(1-p)}{\sqrt{n}}\to^{d} N(0,\tau^2)\,\,\,\,\,\,\,\,......(*)$$ where, $\tau^2=Var(Y_1)+2.Cov(Y_1,Y_2)=p(1-p)-3p^2(1-p)^2$
My solution We have to, $Y_1=X_{2}(1-X_1),\,\,\, Y_2=X_{3}(1-X_2),\,\,\,Y_3=X_{4}(1-X_3),\,\,... $ we see that $Y_1,Y_2,Y_3,Y_4,..$ is a sequence 1-dependent and uniformly bounded r.v's such that
$$\frac{\sigma(T_n)}{n^{\frac{1}{3}}}=\frac{\sqrt{\sum^{n}_{i}Var(Y_i)+\sum^{n-1}_{i}2.Cov(Y_{i},Y_{i+1})}}{n^{\frac{1}{3}}}$$
where, $E(Y_i)=p(1-p),\,\,\,$ $Var(Y_i)=p(1-p)$ and $Cov(Y_i,Y_{i+1})=E(Y_i.Y_{i+1})-(E(Y_i))^2=0-[p(1-p)]^2$, for all $i$. $$\frac{\sigma(T_n)}{n^{\frac{1}{3}}}=\sqrt{\frac{np(1-p)}{n^\frac{2}{3}}+\frac{(n-1)2.Cov(Y_{i},Y_{i+1})}{n^\frac{2}{3}}}\to +\infty$$ this because $\frac{n}{n^\frac{2}{3}}\to +\infty$, and $-1\leq Cov(Y_{i},Y_{i+1})\leq 1$. for all $i$.
For theorem 7.3.1 (A course in probability theory K.L chung). Then
$$\frac{T_n-E(T_n)}{\sqrt{Var(T_n)}}\to^{d} N(0,1)$$
where, $E(T_n)=np(1-p)$ and for $(*)$ $Var(T_n)=n[Var(Y_1+2Cov(Y_1,Y_2)]$
this is my difficulty, I can not prove that: $Var(T_n)=n[Var(Y_1+2Cov(Y_1,Y_2)]$
Thank you for any help.
Here it is, if $n\to \infty$ the result is obtained.